Design of Iron Powder Core Filter Inductor 6

Date:2026-02-02 Categories:Product knowledge Hits:301 From:Guangdong Youfeng Microelectronics Co., Ltd


Calculate N based on the number of ampere turns.
N=NI/I=113/3=37.7, Take 38
Calculate the inductance,
L=N2AL=382×0.07=101.1μH
The following steps are the same as method one and will not be listed again.
Both of the above methods require trial selection. If the trial selection fails, try selecting a larger or smaller specification depending on the situation until the requirements are met.
(3) Method 3: Use the two curves provided by the production unit (see Figure 2 and Figure 3) to identify the required magnetic core.  diode
Design basis: The inductance is 100 μ H, the rated current is 4A, and the inductance is not less than 75 μ H at the rated current.
1) Calculate the energy storage coefficient. When calculating here, it is required that the inductance be the value at rated current, which is 75 μ H.

Calculate the number of ampere turns and inductance.

N=NI/I=130/4=32.5, Take 33

L=332×0.093=101.3μH

This inductance is the value at non rated current.

The next steps are the same as before. Table 4 lists the maximum inductance of four commonly used specifications of iron powder cores at different currents under several μ o conditions. During design, by checking the table, it is possible to determine which type of magnetic core (iron core specifications) to use.  diode

Table 5 lists the magnetic core specifications that can be selected for different rated currents, different% μ o, and not exceeding a certain inductance.

Finally, it should be pointed out that the ampere turns calculated by methods one and two can sometimes be very large and may not be able to fit around in reality. Therefore, it is necessary to verify according to the following formula.  diode

NI=KSoJ(7)

In the formula, K is the window utilization coefficient, taken as 0.4.

So is the window area of the magnetic core, measured in mm2.  diode

J is the current density, measured in A/mm2. Generally, it can be taken as 2-4 A/mm2. The current density increases, allowing for more turns to be wound, but the copper loss increases and the coil temperature rises.



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